Weak continuous convergence of operators

Let $T_n$ and $T$ be linear maps from Banach space $X$ to a Banach space $Y$.

Suppose $T_n$ satisfies $T_nx \to Tx$ (convergence in the $Y$ norm) for all $x \in X$.

Let $x_n \rightharpoonup x$ in $X$ (weak convergence). Under what conditions on $T_n$ and $T$ does this imply that $$T_nx_n \rightharpoonup Tx?$$

Let us assume all maps are bounded if necessary.

What if instead of $x_n \rightharpoonup x$ we hypothesised $x_n \to x$?

migrated from mathoverflow.netApr 21 '15 at 19:50

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• If the maps aren't continuous (bounded), you can't expect (weak) convergence. If we assume boundedness and strong convergence $x_n \to x$, the Banach-Steinhaus theorem implies the result. (Edit: added requirement for strong convergence.) – Daniel Fischer Apr 21 '15 at 22:08
• @DanielFischer: How do you get this from Banach-Steinhaus? I'm not seeing it. – Nate Eldredge Apr 22 '15 at 0:47
• @NateEldredge Good thing you don't see it, because it doesn't hold if we have only weak convergence $x_n \rightharpoonup x$. (Counterexample: $X = \ell^p(\mathbb{N}),\; 1 < p < \infty$ with $x_n = e_n$ and $T_n(y) = y_n$.) I can't reconstruct what mistake made me think otherwise yesterday, unfortunately. Thanks for the heads-up. – Daniel Fischer Apr 22 '15 at 8:37
• @DanielFischer Can you give a hint why it implies the result? – C_Al Apr 23 '15 at 14:59
• Yes, Banach-Steinhaus gives uniform boundedness of the $T_n$, which is equicontinuity, and that gives norm convergence of $T_n x_n$ to $Tx$ if we assume $x_n \to x$. If we assume only $x_n \rightharpoonup x$, we would need norm convergence $T_n \to T$ to be able to deduce $T_n x_n \rightharpoonup Tx$. – Daniel Fischer Apr 23 '15 at 15:50